A 0.1M of na2so4 is dissolved to an extent of 95%,what would be its osmotic pressure at 27 Celsius?

Answers (1)

Dissociation of  Na_2SO_4:

Na_2 SO_4 \rightarrow 2Na^+ + SO_4^{-2}

Hence, one molecule sodium sulphate dissociates to form three molecules.

\Rightarrow n = 3

Now, Van't Hoff's factor, i = 1-\alpha+n\alpha,

where \alpha is degree of dissociation; \alpha = 0.95

\Rightarrow i = 1-0.95+3\times0.95 = 2.9

Now, \pi = iCRT

where, C denotes concentration, R denotes universal gas constant and T is temperature in Kelvin.here, i = 2.9 , C = 0.1\ M, R = 0.082\ and\ T = 27^{\circ}C =300\ K

\Rightarrow \pi = 2.9 \times 0.1 \times 0.082 \times 300K = 7.134\ atm

Therefore, osmotic pressure is 7.134\ atm.

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